Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

I thought if you dropped a line down from a triangle vertex and it formed a right angle on the opposite side then that line bisected the side? So in this case if you know what BD is then you know what DC is?

To figure out whether it holds, why don't you try drawing some extreme figures, say, something like this:

Attachment:

Ques3.jpg [ 3.67 KiB | Viewed 1345 times ]

Will this be true in this case? When will it be true? When the triangle is equilateral, sure. Also when the triangle is isosceles if the equal sides form the angle from which the altitude is dropped.

Don't put your faith in the figure given. It may be just one of the many possibilities or may be somewhat misleading.
_________________

Can I not imagine that, in a right angled triangle if one of the side of a triangle bears x square root 3 as its length can I not imagine that it's a 30-60-90? If not why.

Can I not imagine that, in a right angled triangle if one of the side of a triangle bears x square root 3 as its length can I not imagine that it's a 30-60-90? If not why.

Please help. Thanks Suneel

What do you mean by "imagine"? How does the length of one side define the angles?
_________________

Hey I had the same question as the last poster. How do we know that BD and DC are of the same length? or that angle BAC has been bisected?

They are not equal: \(BD=\frac{6}{\sqrt{3}}\) (from the first statement) and \(AD=6*\sqrt{3}\) (from the second statement).

I think answer is E, since it is not given that BDC points are collinear.

That's not correct.

OG13, page 272: A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2). Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight. You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees. All figures lie in a plane unless otherwise indicated.

OG13, page 150: Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

If AD is 6, and ADC is a right angle, what is the area of equilateral triangle ABC?

Then statement 1 would have been irrelevant and statement 2 would have been incorrect. If side of an equilateral triangle is 12, the altitude would be \((\sqrt{3}/2)*12 = 6*\sqrt{3}\) But AD is given to be 6.
_________________

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

Show Tags

12 Apr 2015, 15:31

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

Show Tags

12 Apr 2015, 17:23

Main take away from this question is: Figures are not drawn to scale.

rest explanations are already given.
_________________

-------------------------------------------------------------------- The Mind is Everything, What we Think we Become. Kudos will encourage many others, like me. Please Give Kudos !! Thanks

what is the area of triangular region [#permalink]

Show Tags

08 Jun 2015, 10:29

If AD is 6 , and ADC is a right angle, what is the area of triangular region ABC? (1) Angle ABD = 60° (2) AC = 12

Attachments

triangleabc.jpg [ 7.64 KiB | Viewed 766 times ]

_________________

The only time you can lose is when you give up. Try hard and you will suceed. Thanks = Kudos. Kudos are appreciated

http://gmatclub.com/forum/rules-for-posting-in-verbal-gmat-forum-134642.html When you post a question Pls. Provide its source & TAG your questions Avoid posting from unreliable sources.

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

Show Tags

25 Oct 2016, 06:46

amitjash wrote:

Attachment:

Triangle.jpg

If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60° (2) AC = 12

I tried not to solve everything as it is a DS question... Area is equal to base x height / 2 AD is height, BC is the base. We need to find the value of BC. 1. since B is 60 degree, A in ABD is 30. we have 1 leg, and we know the property of 30-60-90 right triangle we can find BD. but since we don't know whether AD divides BC in two equal parts, then we can't tell exactly whether we can find BC.

2. AC = 12. We can find DC. we are faced with the same problem - can't find BD.

1+2 we have BD and we have DC. we can find the answer.

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio \(1:\sqrt{3}:2\) --> as AD=6 (larger leg opposite 60 degrees angle) then \(BD=\frac{6}{\sqrt{3}}\) (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

Answer: C.

I thought if you dropped a line down from a triangle vertex and it formed a right angle on the opposite side then that line bisected the side? So in this case if you know what BD is then you know what DC is?

I also assumed a similar thing . But I believe a perpendicular only bisects the angle at the vertex IF the perpendicular bisects the opposite side .. and vice versa . So we cannot assume the perpendicular is an angle / side bisector unless it was explicitly mentioned in the stem

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...